Problem 3 - Algebra and Trigonometry

Question

\(3-10=\) Find the exact value of each expression, if it is defined. $$ \begin{array}{lll}{\text { (a) } \sin ^{-1} 1} & {\text { (b) } \sin ^{-1} \frac{\sqrt{3}}{2}} & {\text { (c) } \sin ^{-1} 2}\end{array} $$

Step-by-Step Solution

Answer

(a) \(\frac{\pi}{2}\), (b) \(\frac{\pi}{3}\), (c) Not defined.

Step 1: Understanding Inverse Sine

The function \( ext{sin}^{-1}(x)\), also known as arcsin, is the inverse of the sine function. It returns the angle \(\theta\) such that \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\) and \(\sin(\theta) = x\). Arcsin is only defined for values of \(x\) in the interval \([-1, 1]\).

Step 2: Evaluate \(\sin^{-1}(1)\)

Find an angle \(\theta\) where \(\sin(\theta) = 1\). For the interval \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\), \(\theta = \frac{\pi}{2}\) produces \(\sin\left(\frac{\pi}{2}\right) = 1\). Therefore, \(\sin^{-1}(1) = \frac{\pi}{2}\).

Step 3: Evaluate \(\sin^{-1}\left( \frac{\sqrt{3}}{2} \right)\)

Find an angle \(\theta\) where \(\sin(\theta) = \frac{\sqrt{3}}{2}\). For the interval \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\), \(\theta = \frac{\pi}{3}\) satisfies this condition since \(\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\). Therefore, \(\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}\).

Step 4: Evaluate \(\sin^{-1}(2)\)

The value \(x = 2\) is outside the domain \([-1, 1]\) for which \(\sin^{-1}(x)\) is defined. Thus, \(\sin^{-1}(2)\) is not defined.

Key Concepts

Arcsin Function

The arcsin function, denoted as \( \sin^{-1}(x) \), is the inverse of the sine function. It helps us find the angle \( \theta \) for which \( \sin(\theta) = x \). This is particularly useful because while the sine function takes an angle as input and gives us a value between \(-1\) and \(1\), the arcsin function reverses this process.
The arcsin function outputs angles in radians, specifically in the range of \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\). Essentially, it links a sine value back to its original angle within this range. This range is crucial because of the sine function's periodic nature, repeating every \(2\pi\) radians.
This ensures every sine value corresponds to a unique arcsin value, eliminating any confusion with repeating cycles.

Domain of Inverse Sine

Understanding the domain of the inverse sine function is key to grasping its limits. The domain of \( \sin^{-1}(x) \) is restricted to the interval \([-1, 1]\).
This means that it's only possible to use arcsin for values within this range. In other words, only sine values between \(-1\) and \(1\) are valid inputs for the function.
Here's why:

• The standard sine function outputs values between \(-1\) and \(1\) for any input angle \(\theta\) within its domain.
• Since the arcsin function is the inverse, it will naturally only operate on this restricted set of outputs.
• Thus, trying to find an arcsin for a value like \(2\) results in "not defined," as \(2\) is outside the permissible interval.

It's important to note this condition whenever you encounter problems involving the arcsin function.

Evaluating Inverse Trigonometric Expressions

Evaluating inverse trigonometric expressions involves finding the specific angle corresponding to a certain sine value.
To evaluate \( \sin^{-1}(x) \), one must:

• Ensure the value \(x\) is within \([-1, 1]\). If \(x\) falls outside this range, like \(\sin^{-1}(2)\), the expression is undefined.
• If \(x\) is valid, identify the angle \(\theta\) which, when given as input to the sine function, results in \(x\). For example, \(\sin^{-1}(1)\) asks, "What angle gives a sine of \(1\)?" The answer is \(\frac{\pi}{2}\).
• Use known angles and their sine values to quickly evaluate expressions. For instance, for \(\sin^{-1}(\frac{\sqrt{3}}{2})\), recognize it as \(\frac{\pi}{3}\).

Comfort with these steps, built on understanding the function and its domain, makes evaluating expressions straightforward.